Small-loss anchor gluing and its iteration obstruction

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The scalar gluing lemma — proved

Let $B\subset\mathbb Z\setminus\{0\}$ be finite, with hereditary dissociation constant $\delta$ and coloring number $q\geq2$. Suppose $w\ne0$, $w\notin B$, and $\chi(B\cup\{w\})=q+1$. In every proper $q$-coloring of $B$, each color class signed-spans $w$: otherwise adding $w$ to that class would give a proper coloring of the extension.

Take the $m$-simplex directions $t_1,\ldots,t_m$, so $\sum_it_i=0$ and any $m-1$ of them are rationally independent, and put $G_m=\bigcup_i Bt_i$. For $m\geq q+1$,

$$\eta(G_m)\geq\delta(1-1/m),\qquad\chi(G_m)=q+1. \tag{1}$$

For extraction, omit a least-populated block of any given subset, and extract within the other blocks. The independent directions make the union dissociated. For the lower coloring bound, any proper $q$-coloring would represent $w$ in each color in every block; summing those representations against $\sum t_i=0$ gives a monochromatic relation. For the upper bound use $q+1$ colors and a proper $q$-palette in every block, arranging that each color is absent from at least one block. Each color then uses at most $m-1$ independent directions.

Exact finite input and improvement

For the checked Grow–Whicher set of Grow–Whicher, Finite unions of quasi-independent sets, the finite input of this route, take $B=E\setminus\{3\}$, $w=3$, $q=2$. The standard-library exact certificate proves

$$\eta(B)=7/13,\qquad\chi(B)=2.$$

A proper partition is $\{5,9,11,13,30,85,243\}$ and $\{4,27,33,81,89,248,253\}$. Hence (1) gives explicit three-color obstructions with

$$\eta(G_m)\geq\frac7{13}(1-1/m)\longrightarrow\frac7{13}>\frac12.$$

This is an improvement of a finite extraction/coloring gap, not unbounded coloring. The finite positive-integer realization lemma applies to these blocks without changing either parameter.

One explicit positive realization is

$$\bigcup_{i=0}^{m-2}4525^i B \ \cup\ (1+4525+\cdots+4525^{m-2})B.$$

Here $\sum_{a\in B}a=1131$; every coefficient in a signed block sum has absolute value at most $2262<4525-1$, so the leading-digit argument prevents additional signed relations. The sign change of the last simplex direction preserves relation supports. For every $m\geq15$, the guaranteed extraction proportion is at least $98/195>1/2$, while the coloring number is exactly three.

Universal anchors do not automatically propagate — proved

For $m\geq q+2$, $G_m$ has no nonzero vector signed-spanned by every color class of every proper $(q+1)$-coloring. In fact the same is true after any linear scalar encoding that preserves the ternary relation system exactly, not just a generic encoding.

Fix a proper $q$-partition $B=P_1\sqcup\cdots\sqcup P_q$. For each $q$-element set of blocks $S$, there is a proper palette coloring in which the new color class is exactly $C_S=\bigcup_{i\in S}P_1t_i$: assign its other $q-1$ old classes to old colors in those blocks, omitting a different old color in each block of $S$, and use only the old colors elsewhere. Every color is absent from a block, so the coloring is proper.

Fix also a $q$-element set $T$ of blocks. Another proper palette coloring has new color $D=\bigcup_{i\in T}P_2t_i$. This set is disjoint from every $C_S$.

If a universal nonzero anchor existed after an exact scalar encoding $L$, choose a signed representation $y$ of it on $D$, and for each $S$ a signed representation $x_S$ on $C_S$. The supports are disjoint, so $L(x_S-y)=0$ is a ternary relation on $G_m$. Exact preservation forces $x_S=y$ already in the vector configuration. This common vector belongs to

$$\bigcap_{|S|=q}\operatorname{span}\{t_i:i\in S\}=\{0\}.$$

For the last equality, choose $q+1\leq m-1$ independent simplex rays and intersect the spans of their $q$-subsets. This contradicts a nonzero scalar anchor. Thus deliberately collapsing coefficient-two differences cannot repair this gluing while preserving all ternary relations.

Boundary case: the Grow source is excluded

For the Grow deletion above, $m=q+1=3$ is also excluded by an exact finite certificate. It proves $U_3(B\otimes S_3)=\{0\}$, even after every exact ternary-preserving scalar encoding. Thus no $m\geq3$ gives a reusable anchor from this source.

A different source is not ruled out. Two sharper tools are available but neither implies anchor survival: the defect estimate below and quadratic-loss primitive-anchor gluing.

Let $r_w^\perp(F)$ be the largest size of a subset of $F$ whose image in the quotient by $\mathbb Qw$ is dissociated, and set $\Delta_w(F)=r(F)-r_w^\perp(F)$. Glue $m$ copies of a vector source by quotienting their direct sum by the single relation $w_1+\cdots+w_m=0$. For layers $F_i$ the elementary bound is

$$r_G\Bigl(\bigcup_i F_i\Bigr) \geq\sum_i r(F_i)-\min_i\Delta_w(F_i). \tag{2}$$

Choose an independent-modulo-$w$ subset in one block and ordinary dissociated subsets in the others. Any cross-block relation would force their block sums to be a common multiple of $w$; the special block rules that out. For scalar sources the special quotient is zero and (2) reduces to whole-block deletion. An invariant making these defects summably small while preserving anchors in later gluings is not established.

Indeed, for $D=B^{\oplus r}$ and diagonal $w=(3,\ldots,3)$, omitting a least-rank coordinate summand before quotienting by $\mathbb Qw$ gives $\Delta_w(F)\leq\min_j r(F_j)\leq r(F)/r$. Its boundary three-copy gluing is $B\otimes S_{3r}$, whose universal three-color anchor set is zero for every $r\geq1$, by the large-arity proof and the new boundary certificate. Small relative defect alone is therefore insufficient.